In the previous section we saw the two tangents from an exterior point. We also saw a solved example. In this section we will see a few more solved examples.
Solved example 32.17
In fig.32.43(a) below, ABC is an isosceles triangle. AC and BC are the equal sides and AB is the base.
All the three sides are tangents to the circle and the tangent points are P, Q and R. Prove that the base AB is bisected at P
Solution:
1. Tangents through P and Q meet at B. So BP = BQ
2. Tangents through Q and R meet at C. So CR = CQ
3. Tangents through P and R meet at A. So AP = AR
4. Given that AC = BC. So we can write:
AR + CR = BQ + CQ
5. But from (2), we have CR = CQ
So (4) becomes: AR = BQ
6. But from (3), we have: AR = AP
Also from (1), we have: BQ = BP
■ So (5) becomes: AP = BP. That means, AB is bisected at P
Solved example 32.18
In fig.32.43(b) above, a circle is touching the side AB of a ΔABC at Q. The circle is also touching the side CA produced at P and CB produced at R. Prove that: CP = Half of the perimeter of ΔABC
Solution:
1. (i) Tangents through P and Q meet at A. So AP = AQ
(ii) Tangents through Q and R meet at B. So BQ = BR
(iii) Tangents through P and R meet at C. So CP = CR
2. The perimeter of ΔABC = AB + BC + CA
• But AB = AQ + BQ
• So perimeter = AQ + BQ + BC + CA
3. From 1(i), we have: AQ = AP
• Also from 1(ii), we have: BQ = BR
• So (2) becomes:
Perimeter = AP + BR + BC + CA
4. Consider the first and last terms in the above equation:
• They are AP and CA
• From the fig., we get: AP + CA = CP
5. So (3) becomes:
• Perimeter = CP + BR + BC
6. Also from the fig., BR + BC = CR
• So (5) becomes:
Perimeter = CP + CR
7. But from 1(iii), we have: CP = CR
• So (6) becomes:
Perimeter = CP + CP = 2 CP
■ Thus we get: CP = Half of perimeter
Solved example 32.19
In fig.32.44(a) below, 3 tangents are drawn at P, Q and R on the circle. The 3 tangents intersect at 3 points to form ΔABC. The tangents at P and R are perpendicular to each other.
Prove that the perimeter of ΔABC is equal to the diameter of the circle.
Solution:
1. From the solved example 32.18 above, we know that AP in fig.32.44(a) = Half the perimeter of ΔABC
2. Now draw the radial lines from R and P. This is shown in fig.b
• Since AR is a tangent, ∠ARO = 90o
• Since AP is a tangent, ∠APO = 90o
3. So in the quadrilateral APOR, 3 angles are 90o. They are:
∠ARO, ∠APO and ∠PAR
• So the fourth angle ∠POR will also be 90o
4. In the quadrilateral APOR, all four angles are 90o
• And adjacent sides AP = AR (∵ they are tangents from an exterior point A)
• So APOR is a square
5. In the square APOR, AP will be equal to OR. So we can write:
• AP = OR
⟹ Half the perimeter of ΔABC = OR
⟹ Full perimeter of ΔABC = 2 × Half perimeter of ΔABC = 2 × OR
But 2 × OR = 2 × radius = diameter of the circle.
■ So we get:
Full perimeter of ΔABC = diameter of the circle
Solved example 32.20
In the fig.32.45(a) below, two circles touch at a point P and a common tangent AB is drawn there
(i) Prove that this common tangent AB bisects the segment RQ on another common tangent CD of the circles (fig.b). Where R and Q are the points of contact.
(ii) Prove that the points of contact P, Q and R form the vertices of a right triangle (fig.c)
Solution:
Part (i):
Let the two tangents AB and CD intersect at S
1. Tangents through P and Q meet at S. So SP = SQ
2. Tangents through P and R meet at S. So SP = SR
3. From (1) and (2) we get: SR = SQ
• So the point S bisects the segment RQ.
■ In other words, the tangent AB bisects the segment RQ
Part (ii):
Consider fig.32.45(d) below.
1. If we draw an arc with S as center and SR as radius, the points P and Q will lie on that arc.
This is because SR = SP = SQ
2. Since R and S lie on the same line, the arc RPQ is a semicircle and RQ is the diameter of that semicircle
3. So for the ∠RPQ, R and Q are the ends of the diameter and P lies on the semicircle. Thus ∠RPQ is a right angle
In the next section, we will see more details about Tangents.
Solved example 32.17
In fig.32.43(a) below, ABC is an isosceles triangle. AC and BC are the equal sides and AB is the base.
 |
| Fig.32.43 |
Solution:
1. Tangents through P and Q meet at B. So BP = BQ
2. Tangents through Q and R meet at C. So CR = CQ
3. Tangents through P and R meet at A. So AP = AR
4. Given that AC = BC. So we can write:
AR + CR = BQ + CQ
5. But from (2), we have CR = CQ
So (4) becomes: AR = BQ
6. But from (3), we have: AR = AP
Also from (1), we have: BQ = BP
■ So (5) becomes: AP = BP. That means, AB is bisected at P
Solved example 32.18
In fig.32.43(b) above, a circle is touching the side AB of a ΔABC at Q. The circle is also touching the side CA produced at P and CB produced at R. Prove that: CP = Half of the perimeter of ΔABC
Solution:
1. (i) Tangents through P and Q meet at A. So AP = AQ
(ii) Tangents through Q and R meet at B. So BQ = BR
(iii) Tangents through P and R meet at C. So CP = CR
2. The perimeter of ΔABC = AB + BC + CA
• But AB = AQ + BQ
• So perimeter = AQ + BQ + BC + CA
3. From 1(i), we have: AQ = AP
• Also from 1(ii), we have: BQ = BR
• So (2) becomes:
Perimeter = AP + BR + BC + CA
4. Consider the first and last terms in the above equation:
• They are AP and CA
• From the fig., we get: AP + CA = CP
5. So (3) becomes:
• Perimeter = CP + BR + BC
6. Also from the fig., BR + BC = CR
• So (5) becomes:
Perimeter = CP + CR
7. But from 1(iii), we have: CP = CR
• So (6) becomes:
Perimeter = CP + CP = 2 CP
■ Thus we get: CP = Half of perimeter
Solved example 32.19
In fig.32.44(a) below, 3 tangents are drawn at P, Q and R on the circle. The 3 tangents intersect at 3 points to form ΔABC. The tangents at P and R are perpendicular to each other.
 |
| Fig.32.44 |
Solution:
1. From the solved example 32.18 above, we know that AP in fig.32.44(a) = Half the perimeter of ΔABC
2. Now draw the radial lines from R and P. This is shown in fig.b
• Since AR is a tangent, ∠ARO = 90o
• Since AP is a tangent, ∠APO = 90o
3. So in the quadrilateral APOR, 3 angles are 90o. They are:
∠ARO, ∠APO and ∠PAR
• So the fourth angle ∠POR will also be 90o
4. In the quadrilateral APOR, all four angles are 90o
• And adjacent sides AP = AR (∵ they are tangents from an exterior point A)
• So APOR is a square
5. In the square APOR, AP will be equal to OR. So we can write:
• AP = OR
⟹ Half the perimeter of ΔABC = OR
⟹ Full perimeter of ΔABC = 2 × Half perimeter of ΔABC = 2 × OR
But 2 × OR = 2 × radius = diameter of the circle.
■ So we get:
Full perimeter of ΔABC = diameter of the circle
Solved example 32.20
In the fig.32.45(a) below, two circles touch at a point P and a common tangent AB is drawn there
 |
| Fig.32.45 |
(ii) Prove that the points of contact P, Q and R form the vertices of a right triangle (fig.c)
Solution:
Part (i):
Let the two tangents AB and CD intersect at S
1. Tangents through P and Q meet at S. So SP = SQ
2. Tangents through P and R meet at S. So SP = SR
3. From (1) and (2) we get: SR = SQ
• So the point S bisects the segment RQ.
■ In other words, the tangent AB bisects the segment RQ
Part (ii):
Consider fig.32.45(d) below.
 |
| Fig.32.45(d) |
This is because SR = SP = SQ
2. Since R and S lie on the same line, the arc RPQ is a semicircle and RQ is the diameter of that semicircle
3. So for the ∠RPQ, R and Q are the ends of the diameter and P lies on the semicircle. Thus ∠RPQ is a right angle
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